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Kirchhoff’s rules Rules for resistors in series and parallel very useful, but: Not all resistor networks can be reduced to simple combinations of series and parallel circuits such as this one How to find this equivalent resistance in most general cases Kirchhoff’s rules http://en.wikipedia.org/wiki/Gustav_Kirchhoff 2 fundamental laws of nature allow to deduce Kirchhoff’s rules Conservation of charge I n n 0 Kirchhoff’s junction rule At any junction* in a circuit the sum of the currents into any junction is zero *Junction (also called node): is a point in a circuit where three or more conductors meet example for a junction not a junction Kirchhoff’s junction rule and the flowing fluid analogy I2 I1 I3=I1+I2 Note: obviously we have to carefully identify the signs of the partial currents when applying the junction rule I1+I2+I3=0 Sign convention needed: We count current flowing into a junction (I1,2 in our example above) positive and those flowing out (I3 in our example above) negative Electrostatic force is conservative Potential energy and potential are unique functions of position b Vab E ( r )d r dV Kirchhoff’s loop rule E ( r )d r 0 a V n n 0 the sum of the potential differences in any loop (including those with emfs) is zero Sign convention and examples We first investigate an elementary simple network R 1 Let’s start by labeling all quantities I 2 Let’s choose a direction for the assumed + current, e.g., clockwise 3 In this simple example there are no junctions We only need the loop rule here r 4 Consider yourself a positive charge traveling the loop E + -when we flow with the current through R we loose potential energy V1 IR -when we flow through the source of emf from – to + we gain potential energy V2 E -when we flow with the current through r we loose potential energy V1 V2 V3 IR E Ir 0 IR E Ir as seen before V3 Ir Can we travel in the opposite direction and get the same result R We keep the current direction but travel against the current I + -when we flow against the current through r we win potential energy V1 Ir r E -when we flow through the source of emf from + to - we loose potential energyV2 E -when we flow against the current through R we win potential energy V1 IR V1 V2 V3 Ir E + IR 0 Now an example that involves the junction and loop rule together Example: charging a battery from our textbook Young and Freedman The direction of this current is our choice Travel-directions we choose to go through the loops Goal: -determine unknown emf of the run-down battery -determine internal resistance r of the 12V power supply -determine the unknown current I We will need 3 equations for the 3 unknowns We apply the junction rule to point a 1 Junction rule at point a: +2A+1A-I=0 I=3A 2 Loop rule for outer loop 1: 12V 3 A r 2 A 3 0 3 Loop rule for loop 2: E 1V 6V 0 r 2 E 5V The minus signs says that the polarity of the battery is opposite to what we assumed in the figure It better is opposite, because who would try to jumpstart a car by connecting terminal of different sign, not a good idea! 4 Loop rule for loop 3 can be used to check : 12V Ir 1A 1 E 12V 6V 1V 5V 0 Example of a complex network which cannot be represented in terms of series and parallel combinations I6 I4 I5 direction of I3 is an assumption which we may or may not confirm throughout the calculation Goal: Finding all 6 unknown currents and the equivalent resistance 1 Junction rule at c: I 6 I1 I 2 0 I6 I1 I 2 2 Junction rule at a: I1 I 4 I3 0 I 4 I1 I 3 3 Junction rule at b: I 2 I5 I 3 0 I5 I 2 I 3 We could apply the junction rule at d No new eq. just confirming I 6 I1 I 2 We need 3 additional equation 1 loop1: applying the loop rule for 3 loops 13V I1 1 I1 I3 1 0 2 loop2: 13V I 2 1 I 2 I3 2 0 3 loop3: I1 1 I 3 1 I 2 1 0 We could apply the loop rule for the remaining loop From 3: I1 I 2 I 3 into 1: No new eq. just confirming 13V I 2 1 I 3 1 I 2 2 I 3 1 0 13V 2 I 2 1 3I 3 1 0 Together with eq. 2 we get 13V 2I 2 1 3I3 1 0 39V 6I 2 1 9I 3 1 0 13V 3I 2 1 2I3 1 0 26V 6I 2 1 4I 3 1 0 13V 13I 3 1 0 I 3 1A minus sign indicates that I3 flows opposite to our assumption in the figure I2 5A I1 6 A Total current through the network I 6 I1 I 2 11A Equivalent resistance Req 13V 1.18 11A Check the result with remaining loop: I 3 1 I 3 I 2 2 I1 I 3 1 0 1V 8V 7V 0 Clicker question What happens to the brightness of bulbs A and B when bulb C is removed from this circuit? For simplicity let’s assume A,B and C are identical 1) No change in A, B gets brighter 2) A and B get brighter 3) A and B get dimmer 4) No change in A, B gets dimmer 5) A gets dimmer, B gets brighter First case with bulbs A,B and C installed: The emitted light intensity depends on the dissipated power of each bulk We calculate total resistance R2 3 Rtot R R 2R 2 I tot 2 2 2V 4V PA R 9R 3R 2V 3R 2V V V 2 PB Pc V 3 3R 9 R Check: total dissipated power 2V 2 4V 2 V 2 2V 2 Ptot PA PB PC 2 3R 9R 9 R 3R Second case with C removed: Rtot 2 R (total resistance increased) I tot V 2R (total current decreased) (voltage drop across A decreased) 2 2 V V PA R 2 R 4R (bulb A gets dimmer) VV V2 PB V 2 2R 4R (bulb B gets brighter) (voltage drop across B increases) V2 Ptot 2R (reduced because current goes down while V=const)